Question

How do you rationalize the denominator `sqrt2/(2-sqrt2)`?

Answer 1

Multiply by `1` in the form: `(2+sqrt2)/(2+sqrt2)`.
(This doesn't change the value of the number, it only changes how the number is written.)

`sqrt2/(2-sqrt2)=sqrt2/(2-sqrt2) * (2+sqrt2)/(2+sqrt2)=(sqrt2(2+sqrt2))/((2-sqrt2)(2+sqrt2))`. So,

`sqrt2/(2-sqrt2) = (2sqrt2 + (sqrt2)^2)/(4-(sqrt2)^2) = (2sqrt2 + 2)/(4-2)`

`sqrt2/(2-sqrt2) = (2sqrt2 + 2)/2 = (2(1+sqrt2))/2 = 1+ sqrt2`

Why does it work?

We can verify that `(a-b)(a+b)=a^2-b^2` for any numbers, `a` and `b`.

So we can also verify that `(a-sqrtc)(a+sqrtc)=a^2 -c`. (no square roots).

By the way, we can also verify that `(sqrta-sqrtb)(sqrta+sqrtb)` has no square roots.